Question

我在mysql中有一个学生月度表我在records_day栏中记录了每月1,2,3,4,5,6..31的每一天。与数学，科学，英语等主题明智。等

月表如：

subject          recorded_day
Math              1
science           1
English           1
Math              2
science           2
English           2
Math              4
science           4
English           4
Math              7
science           8
English           9

如何编写查询以检索在课堂上不存在的数据分离的主题学生。我想要检索录制的日期哪个学生不在主题明确的班级

输出

subject       recorded_day not present
Math        3,5,6,8,9,10...31
science     3,5,6,7,9......31
English     3,5,6,7,8,10,...31

Answer 1

试试这个：

SELECT t.Subject, GROUP_CONCAT(t.row order by t.row) 
FROM
(
  SELECT row, subject
  FROM
  (
    SELECT  @row := @row + 1 as row FROM 
    (select 0 union all select 1 union all select 3 union all 
     select 4 union all select 5 union all select 6 ) t,
    (select 0 union all select 1 union all select 3 union all 
     select 4 union all select 5 union all select 6 ) t2,
    (SELECT @row:=0) AS t3
  ) as sub, 
  (
    SELECT 'Math' AS subject UNION ALL SELECT 'science'
    UNION ALL SELECT 'English') AS subjects
  WHERE row <= 31 
) AS t
LEFT JOIN Table1 AS t1 
         ON t.row = t1.recorded_day AND t.subject = t1.subject
WHERE t1.recorded_day IS NULL
  GROUP BY t.subject;

以下是它的工作原理：

生成1到31的序列号。
为每个主题获取此列表的笛卡尔积。
LEFT JOIN原始表格，以获取那些不存在的日子。
将GROUP_CONCAT与GROUP BY一起使用。

SQL Fiddle demo

结果：

| Subject |                                          NotPresent 
|---------|-----------------------------------------------------------------------------|
| English | 3,5,6,7,8,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31 |
|    Math | 3,5,6,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31 |
| science | 3,5,6,7,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31 |

Answer 2

您需要创建一个包含几天和所有主题的表格（month_days）：

days                 subject 
1                     Math
2                     Math 
3                     Math 
.
.
.
31(or 30, 28, 29)     English
1                     English
2                     English 
3                     English 
.
.
.
31(or 30, 28, 29)     English
.
.

SELECT subject, GROUP_CONCAT(recorded_day SEPARATOR ',') 
FROM month_days m_d
LEFT JOIN table t on t.recorded_day=m_d.days and t.subject=m_d.subject
WHERE t.recorded_day is null
GROUP BY subject;

学生不会出现明智的学生记录的主题

2 个答案: